1

I am asking this question as I have not seen an answer to the question (that I can understand) as of yet (Credit for original question: Series expansion of infinite series raised to the $n$th power)
Can anyone explain how you can do it for an infinite $$\left( \sum_{n\in \mathbb N} \ f(x,n) \right)^n$$

Sorry for the bad explanation, linking and other stuff, I am not familiar with this site, but please explain the answer to this question as I am not quite familiar with infinite sums yet

Sebastiano
  • 7,649
thecodeadd
  • 13
  • 3

1 Answers1

1

Your example has three n's, the outer has to be another symbol

By definition (and if all terms exist and are unique)

$$\left( \sum_{n\in \mathbb N} \ f(x,n) \right)^m=\prod _{k=1}^m\left( \sum_{n_k\in \mathbb N} \ f(x,n_k) \right) $$

Any further simplification or expansion needs an ordering of all monomials in $f(x,k_j)$. This is simple by completely symmetric product structure of the f(x,n)f(x,m), eg powers, exponentials or the famous finite algebras.

For the classical complex Taylor series, one collects all products with constant total exponent

$$\prod _{k=1}^m\left( \sum_{n_k\in \mathbb N} \ a_{n_k} \ x^{n_k} \right) = \sum_{n\in \mathbb N} x^n \quad \left(\sum_{\sum _j k_j=n}\prod a_{k_j}\right) $$

Exercise: Power of Exponential series

$$\prod _{k=1}^m\left( \sum_{n_k\in \mathbb N} \ \frac{\ x^{n_k}}{n_k!} \right) = \sum_{n\in \mathbb N} x^n \quad \left(\sum_{\sum _j k_j=n}\prod \frac{1}{k_j!}\right) = \sum_{n\in \mathbb N} \frac{(m x)^n}{n!} $$

Roland F
  • 2,098
  • Clarification about the exercise, how did you get sum of ((mx)^n)/(n!) from the sum. Thank you for your answer and explanation! – thecodeadd Jan 12 '24 at 01:56
  • The central point in the proof of the addition theorem of exponentials, trigonometrics, elliptic functions, in general, is the binomial theorem: $$ (\sum a_i)^k = \text{pick an $a_i$ from all brackets, multiply and add},$$ generalization of $$(a+\sum_b_i)^2=a^2+ 2 a \sum b_i +(\sum b_i)^2$$ – Roland F Jan 12 '24 at 06:05
  • Another question about the notation (Sorry about the amount of questions, I was not able to edit my original comment further) but I am used to seeing bounds on the summation and product functions, what does it mean this this case (I understand n belongs to subset of all natural numbers but I do not understand sum of j where k_j = n and the product function has no bounds written) Could you please explain the bounds for further clarification? Thank you for the time and effort in explaining this, I appreciate the help! – thecodeadd Jan 12 '24 at 09:30
  • Linear ordered sums, integrals are a central in sums of infinitely many objects in one dimension, in cases, where the sum is not stable with respect to reordering. On the other hand, for sums, that converge absolutely, order does not play a role and domains of summation and integration can simply denoted by the domain. This notation has the great advance, that set manipulations can be done without using maps explicitely. Famous examples are Lebesgue integrals and Fubini decompositions of sets in $R^n$. For finite multiple sums the set decompostion is easy, renumbering is error prone. – Roland F Jan 12 '24 at 12:13
  • Oh ok, so that would imply that the product operator has the sam bounds as originally, except its k_j = 0 instead of k=0, and the infinite sum of k_j is equal to n. Please correct me if I'm wrong. Thank you for the explanation of this notation! – thecodeadd Jan 12 '24 at 22:58